Question

The locus of mid-point of line segment intercepted between real and imaginary axes by the line $a\overline{z}+\overline{a}z+b=0$ where $b$ is a real parameter and $a$ is a fixed complex number with non- zero real and imaginary parts, is

• A. $az+\overline{a}z=0$
• B. $a\overline{z}+\overline{a}z=0$
• C. $az+\overline{az}=0$
• D. $az-\overline{a}z=0$

Answer 1

The correct option is C $az+\overline{az}=0$

Given equation of line is
$a\overline{z}+\overline{a}z+b=0,\forall b\in R$
Let $PQ$ be the segment intercepted between the axes. For real intercept $z_R$.
$z_R=\overline{z_R}$
$\Rightarrow z_R(a+\overline{a})+b=0$
$\Rightarrow z_R=\frac{-b}{(a+\overline{a})}$
For imaginary intercept $z_I$,
$z_I+\overline{z_I}=0$
$\Rightarrow z_I(\overline{a}-a)+b=0$
$\Rightarrow z_1=-\frac{b}{\overline{a}-a}$
Let locus of mid- point be $z$ ,
$z=\frac{z_R+z_I}{2}$
$=\frac{-b}{2}[\frac{1}{\overline{a}+a}+\frac{1}{\overline{a}-a}]$
$=\frac{\overline{a}b}{(a+\overline{a})(a-\overline{a})}$
$=\frac{\overline{a}b}{a^2-(\overline{a}{)}^2}$
$\Rightarrow z\frac{a^2-(\overline{a}{)}^2}{\overline{a}}=b$
As $b$ is real parameter,
$\Rightarrow z\frac{a^2-(\overline{a}{)}^2}{\overline{a}}=\overline{z}\frac{(\overline{a}{)}^2-(a{)}^2}{a}$
$\Rightarrow az+\overline{az}=0$